An aromatic compound A ( C ₉ H ₁₂ ) is formed by passing alkyne B through a red-hot iron tube. Compound A on nitration yields a single mononitro derivative. Alkyne B can be prepared by the reaction of a dihalide C with excess NaNH ₂ . Furthermore, B on hydration with dil. H ₂ SO ₄ and HgSO ₄ gives compound D. Identify the correct pair of compounds C and D.

Step-by-step solution

The aromatic compound A ( C ₉ H ₁₂ ) is formed by the trimerization of alkyne B. This indicates that B is a 3-carbon alkyne, which is propyne ( CH ₃ C CH ). Trimerization of propyne in a red-hot iron tube yields 1,3,5-trimethylbenzene (mesitylene) as compound A. Mesitylene has all three of its ring hydrogen atoms in equivalent positions, so it yields a single mononitro derivative upon nitration, confirming our deduction. Alkyne B (propyne) is prepared by the double dehydrohalogenation of a dihalide C using excess N